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frutty [35]
3 years ago
10

"Atoms of magnesium and oxygen form ions and ionic bonds "The difference is that instead of having each atom gain or lose one el

ectron, each atom of one of these elements loses two electrons, and each atom of the other element gains two electrons. Describe the process that leads to the formation of the ionic bond between magnesium and oxygen atoms in magnesium oxide.

Chemistry
1 answer:
Tanzania [10]3 years ago
5 0

Explanation:

Atomic number of magnesium is 12 and its electronic distribution is 2, 8, 2. To attain noble gas configuration it is necessary for the magnesium atom to lose two valence electrons and therefore, it forms Mg^{2+} ions.

On the other hand, atomic number of oxygen atom is 8 and its electronic distribution is 2, 6. To attain noble gas configuration it needs to gain two electrons. And, on gaining the electrons it forms O^{2-} ions.

Hence, when both magnesium and oxygen ions chemically combine with each other then it forms the compound magnesium oxide (MgO).

This is because magnesium transfer its two valence electrons to the oxygen atom and due to the formation of opposite charges on these atoms they get attracted towards each other.  

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4 years ago
Calculate the pH for each of the cases in the titration of 25.0 mL of 0.180 M pyridine, C 5 H 5 N ( aq ) with 0.180 M HBr ( aq )
Oxana [17]

Answer:

Explanations

Calculate the pH for each of the following cases in thetitration of 25.0 mL of 0.100 M pyridine, C5H5N(aq) with 0.100 MHBr(aq):

a.Before and addition of HBr

b.After addition of 12.5ml HBr

c.After addition of 15ml HBr

d.After addition of 25ml HBr

e.After addition of 33ml HBr

SOLUTION ;;;

Kb of pyridine =1.5*10^-9

a)let the dissociation be x.so,

kb=x^2/(0.1-x)

or 1.5*10^-9=x^2/(0.1-x)

or x=1.225*10^-5

so,

[OH-]=1.225*10^-5

so,

pOH=-log([OH-])

so pH=9.088

b)now this will effectively behave as a buffer

pKb=8.82

so pOH=pKb+log(salt/acid)

=8.82+log((12.5*0.1)/(25*0.1-12.5*0.1))

=8.82

so pH=14-pOH

=5.18

c)again using the same equation as the above,

pOH=pKb+log(salt/acid)

=8.82+log((15*0.1)/(25*0.1-15*0.1))

=9

so pH=14-9

=5

d)now the base is completely neutralised.so,

concentration of the salt formed=0.1/2

=0.05 M

so,

pH=7-0.5pKb-0.5log(C)

=7-0.5*8.82-0.5*log(0.05)

=3.24

e)concentration of H+=(33*0.1-25*0.1)/(33+25)

=0.01379

so pH=-log(0.01379)

=1.86

4 0
3 years ago
Read 2 more answers
A cell consists of a gold wire and a saturated calomel electrode (S.C.E.) in a 0.150 M AuNO 3 solution at 25 °C. The gold wire i
almond37 [142]

The half reaction that occurs at the Au electrode is  1.64

<u>Explanation:</u>

Half cell reaction at 'Au' electrode

We have the equation,

Au + (aq) + e-  ---> Au(s)

Given the concenration of AuNO3=0.150 M

[Au +] =0.150 m

From the equation,

Au + (aq) + e-  ---> Au(s)

Standard electric potential= Eo= 1.69 volt

Solving the problem using the Nerst equation

E cell= E0 cell - 2.303 RT/ nF  log Qc

Where,

T = 298 K

n= no of electron lost or gained

F= faraday's constant= 965000/mole

R= universal constant= 8.314 J/ K/ Mole

Substitue the values we get

E cell = 1.69 volt - 0.05 g/n log (1/0.150M)

E cell = 1.69 volt - 0.05 g/1  0.824

E cell= 1.64

The half reaction that occurs at the Au electrode is  1.64

<u />

6 0
3 years ago
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